$sin (2 \alpha) =2\cdot sin \alpha \cdot cos \alpha \\cos(2 \alpha )=1-2sin^2 \alpha \\-------------------- \\cos2x-sin2x=1-2sin2x\\cos2x+sin2x=1\\1-2sin^2x+2sinx\cdot cosx=1\\-2sinx(sin x-cosx)=0\\ -2sinx=0\ \ \ \ or\ \ \ \ sinx-cosx=0\\\\1)\ -2sinx=0\ \ \Leftrightarrow\ \ sinx=0\ \ \Leftrightarrow\ \ x_1=k \pi \ \ and\ \ k\in I\\2)\ \ sinx-cosx=0\ \ \Leftrightarrow\ \ sinx=cosx\ /:cosx\ \ \ \ \ \wedge\ \ \ \ cos x \neq 0\\tanx=1\ \ \Leftrightarrow\ \ x_2= \frac{ \pi }{4} +k \pi \ \ and\ \ k\in I$

3 0

3 years ago

FIRST ANSWER GETS BRAINLIEST

Anarel [89]

Answer:

4/15

Step-by-step explanation:

You would do it as follows

I-8/15I reduce the fraction by 2

I-4/25I the absolute fraction is always positive

So the solution is

4/25

Alternative forms are:

0.16 or (2/5)^2 (Just incase)

5 0

2 years ago

Read 2 more answers